Explain the numbers - imaging circle vs sensor size

Public admission time. :(
I work in finance but failed arithmetic at school (no wonder we had a GFC). I am not talking about algebra, calculus, etc just arithmetic.:eek:
I understand the sensor size and crop factor on SLR sensors but with compact and bridge cameras we get sensor sizes like 4/3”, 2/3”, ½”, etc.
I also understand that these smaller sensors are 5, 6 or 7 times approximately smaller than a FF sensor. I simply divide the smaller size lens focal length into the FF equivalent provided in the specs to get his result.
BUT
How do the fractions in the second paragraph above translate to “x – times” smaller sensor?
Ps. Please keep my maths problem a secret.:)

Hi Peter,

Does this help?

Hi Colin,

I have read that tutorial a number of times (and all tutorials for that matter and find them extremely informative) and understand the concept but I cannot work out how a 4/3" relates to x-time smaller sensor.

I suppose my problem with anything to do with maths or numbers has always been that I have never been able to explain what I need to know to clarify the problem in my mind.

What I am trying to understand is how for example 4/3” turns into x – time smaller lens. What does 4/3” actually represent? Is it a size, a ratio, a state of mind? I do not understand what the fraction, number, or whatever it is actually represents.

I understand the 1.5 or 1.6 crop factors in Nikon and Canon sensors but as the sensors get smaller thay change the way the size of the sensor is stated and it is this new number/fraction, etc I do not understand what is means.

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Originally Posted by Peter Ryan

I have read that tutorial a number of times (and all tutorials for that matter and find them extremely informative) and understand the concept but I cannot work out how a 4/3" relates to x-time smaller sensor.

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In general, this imaging circle is a little larger than the distance across the camera sensor's diagonal, but there's no strict correlation; for each sensor size specification you still need to look up its exact dimensions. It all depends on how much a given camera format chooses to extend its sensor into the lesser quality (outer) regions of this imaging circle. The sensor size naming convention is a remnant from a (relatively) long time ago when it was used for TV camera tubes. Just think of it as a rough indicator of sensor size, that's all.

There's also a visual depiction of the sensor vs imaging circle at the top of this tutorial:
Tilt Shift Lenses: Perspective Control

Sometimes the best kind of math is when you can avoid it entirely...

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Sometimes the best kind of math is when you can avoid it entirely...

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Thanks Sean (and also Colin) for following through on my confusion. I have managed to survive life without general maths but for whatever reason I do understand financial maths - luckily for my clients.

I appreciate the clarification and please keep the tutorials comming - I love reading and re-reading them.

Look at this Wiki article:

http://en.wikipedia.org/wiki/Image_sensor_format

About halfway down is a Table of Sensor Sizes with all the math worked out.

Peter,

Take a look at this Wikipedia page:

Image_sensor_format

To your question

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What does 4/3” actually represent? Is it a size, a ratio, a state of mind?

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It is explained:

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The sensor sizes of many compact digital cameras are expressed in terms of the
non-standardized "inch" system, as approximately 1.5 times the length of the diagonal of
the sensor. This goes back to the way image sizes of early video cameras were expressed
in terms of the outside diameter of the glass envelope of the video camera tube. Instead of
"formats", these sensor sizes are often called types, as in "1/2-inch-type CCD."

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What does 4/3” actually represent?

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Now, pull out your financial calculator to do some simple math.

Turn the fraction into a real number:

4 / 3 = 1.3333"

Divide this by 1.568, the number derrived from the table of imager sizes:

1.3333 / 1.5678 = .8504" - This is the diagonal dimension of a 4/3" imager

Since the aspect ratio is commonly 4:3, we can find the length and width by using the pythagorean theorem from geometry:

Diagonal = SQRT( Width^2 + Height^2 )

and we solve this algibraic formula to find the value of x:

.8504 = SQRT( ( 4 * x )^2 + ( 3 * x )^2 )

Maxima, an algebraic/calculus processor that I use, simplified the whole formula to:

.8504 = 5 * x

where x turns out to be 0.1700"

Therefore, the imager width is 4 * 0.1700" = 0.680" and the height is 3 * 0.1700" = 0.510"

I hope this answers your question.

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Originally Posted by Steaphany

Now, pull out your financial calculator to do some simple math.

Turn the fraction into a real number:

4 / 3 = 1.3333"

Divide this by 1.568, the number derrived from the table of imager sizes:

1.3333 / 1.5678 = .8504" - This is the diagonal dimension of a 4/3" imager

Since the aspect ratio is commonly 4:3, we can find the length and width by using the pythagorean theorem from geometry:

Diagonal = SQRT( Width^2 + Height^2 )

and we solve this algibraic formula to find the value of x:

.8504 = SQRT( ( 4 * x )^2 + ( 3 * x )^2 )

Maxima, an algebraic/calculus processor that I use, simplified the whole formula to:

.8504 = 5 * x

where x turns out to be 0.1700"

Therefore, the imager width is 4 * 0.1700" = 0.680" and the height is 3 * 0.1700" = 0.510"

I hope this answers your question.

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I think I'm impressed. May I ask what you do for a living, Steaphany?

Hi, Peter;

I'm curious as to the back story for the question. It's a very legitimate question, of course, but if you explain what you're trying to get at, it will help, I think. I'm assuming you don't plan to take the sensor out of a camera and mount it elsewhere, so the physical sensor sizes don't directly matter much to you.

As described in the tutorials, a sensor is a grid of pixels that (as we know) is illuminated by light admitted through the lens. The combination of megapixels and sensor size determines the size of the pixels, which is important for noise, because each sensor pixel is a bucket that accepts photons, and for any given situation, a fixed number of photons are available for a unit area. The light coming through the lens will blur at the sensor due to diffraction, so even with the same number of megapixels, a physically smaller sensor will have problems at smaller apertures.

I think the links to the tutorials and to wikipedia will make for easy calculation of physical sensor size and pixel size for a given sensor. The question is what to do with that information.

One thing to clarify: 4:3 is the aspect ratio for the Four Thirds system, and holds for 4/3, 1/1.7, 1/1.8, and 1/2.5 sensors. Full frame and "cropped" (1.5 and 1.6 crop) DSLR sensors are 3:2 aspect ratio. That is, in landscape orientation, the width to the height is a 3 to 2 ratio.

Cheers,
Rick

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Originally Posted by carregwen

I think I'm impressed. May I ask what you do for a living, Steaphany?

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Rob,

This is my corporate web site:

http://kitsune-eng.com/

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Originally Posted by rick55

The question is what to do with that information.

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Actually there are things a photographer can do once they know the width and height of their camera's imager.

If you are working with a compact digital camera, then calculate the width and height as shown above. Since pretty much all camera's have their focal length or range of possible focal lengths in milimeters, you'll need to convert the inch values.

milimeters = inches * 25.4

For cameras with larger imagers, the specifications probably provide milimeter width and height.

This will allow you to calculate the angle of view from the imager's dimensions and the lens's true focal length. (This is one area where you can forget about the crop factor)

Angle of View = 2 * Arc Tangent( Imager Dimension / ( 2 * Focal Length ))

where Imager Dimension can be width, height, or the diagonal

I have a spread sheet which matches camera body to compatible lenses in my collection and calculates the angle of view. I find knowing these values handy when I'm doing astronomical photography where everything is measured in angles as well as when I want a specific angle of view to achieve a particular look to a photograph.

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Originally Posted by Steaphany

Actually there are things a photographer can do once they know the width and height of their camera's imager.

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Excellent point: I said that badly. Instead of saying, "The question is what to do with the information," I should have said, "The question is what you want to do with the information." Are you worried about noise, DOF, zoom?

Cheers,
Rick

I very much appreciate everyone’s input here.
The reason behind the question Rick was simply when reading the tutorial I could see the comparable sensor sizes drawn in the diagram but could not understand the logic behind the numbers.
Ben, I thank you for the Wiki reference and to Steaphany I also say thanks for the explanation but my issues with maths go way back and if you put “x” into any equation I am lost. I will take your word on the example shown.
Having said that I do learn and understand from reading ‘the story’ so all this discussion and the reference articles have helped me understand that the numbers in the tutorial diagram of the sensor sizes, while showing convention, do not in themselves have a logical progression in their expression with
1. 5 times being a crop factor, 4/3 being an aspect ratio and the other being older naming conventions from early video cameras and referred to as ‘types’. I understand there are ways to convert these (for those less challenged than I) but for someone like me when these are all shown on one diagram they are not directly comparable measurements and I try to understand.
As long as I understand that and read what the differences are I can live with it.
Hopefully my understanding is correct. Thanks to everyone.

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Ben, I thank you for the Wiki reference and to Steaphany I also say thanks for the explanation but my issues with maths go way back and if you put “x” into any equation I am lost. I will take your word on the example shown.

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'x' here is just a number and all that is being used is school arithmetic and Pythagorus known to Babylonians proved by Euclid; 'a triangle (x ,y ,z) is right angled if and only if x^2 + y^2 = z^2.
The impressive bit is the program that recognised the (3,4,5) triangle where 9+16=25 and Steaphany's superior knowledge of photography.
But I don't know if anything is gained from knowing all this stuff unless your an engineer.

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Originally Posted by Peter Ryan

...How do the fractions in the second paragraph above translate to “x – times” smaller sensor?

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Errr... they don't, directly, although as everyone's describing above, it does describe the sensor dimensions in a roundabout way. It's a video tube standard "type" designation, not an actual dimension of the sensor. To quote from the dpreview article on this:

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The 'Type' designation given to today's CCD sensors harks back to a set of standard sizes given to TV camera tubes in the 50's. These sizes were typically 1/2", 2/3" etc. The size designation does not define the diagonal of the sensor area but rather the outer diameter of the long glass envelope of the tube. Engineers soon discovered that for various reasons the usable area of this imaging plane was approximately two thirds of the designated size. This designation has clearly stuck (although it should have been thrown out long ago). There appears to be no specific mathematical relationship between the diameter of the imaging circle and the sensor size, although it is always roughly two thirds.

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